PROBLEM 053 · Expert · The current-mirror OTA
Twice as Nice
A current-mirror OTA with a gain of two
The problem
Build a current-mirror OTA. An NMOS pair (gates “inp” and “inn”, tail Current source) has a diode-connected PMOS load on each drain (M3 on the “inp” side, M4 on the “inn” side). Each diode is mirrored by a PMOS with twice its width: M5 copies M3 and feeds “out” directly; M6 copies M4 and feeds a diode-connected NMOS (M7), which an NMOS (M8) of the same size mirrors to pull current from “out”. The output is the junction of M5’s and M8’s drains.
The output current is K·(I3 − I4) with K = 2: the effective transconductance is K·gm1, twice that of the five-transistor OTA from the same pair. With the output resistance of only two transistors (ro5 ∥ ro8) the gain is K·gm1·Rout, and the bandwidth K·gm1/(2π·C_L) is twice as high. The price is two more mirrors, which add poles of their own.
The test bench
- Supply: VDD = 1.8 V.
- The test bench adds a 1 pF load capacitor on “out”, and a 1 GΩ feedback resistor from “out” to “inn” (with a 1 F capacitor from “inn” to a quiet source) so the amplifier biases itself but is open-loop for AC.
- It applies the AC stimulus to “inp” only.
Numbers to hit
- DC gain at least 36 dB
- Unity-gain bandwidth at least 25 MHz
- Phase margin at least 45 °
- Output DC level between 0.5 V and 1.3 V
- Power at most 250 µW
- Every transistor in saturation
Design it yourself
Draw the circuit on a schematic canvas, size every transistor, and see it simulated in your browser with ngspice and the SkyWater SKY130 models. Every transistor lights up green, gold or red by operating region as you move the sliders.
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